studentkakraaLv1

22 Aug 2022

Solve the system of equations.

 

x+4 y-z=14 

x+5 y+z=27 

x-4 y+5 z=-2

 

Select the correct choice below and fill in any answer boxes within your choice.

A. The one solution is x = ___, y = ___, and z =___. (Simplify your answers.)

B. There are infinitely many solutions. If z is allowed to be any real number, then x =___ and y = ___ (Type expressions using z as the variable.)

C. There is no solution.

 

View an example I All parts showing

Solve the system of equations.

 

x+6 y-z=-19 

x+7 y+z=-18 

x-5 y+7 z=30

 

Use the left-to-right elimination method to solve this system of equations. This method reduces the system to an equivalent system in echelon form shown below, where a, b, c, d, e, and f are constants.

 

x + a y + bz = d 

         y + cz = e 

                 z = f

 

Note that the first equation, x + 6y - z = -19, is already in the form of the first row in echelon form.

 

The operations shown helom when perisemed on a system of equations, result in an equivalent system

1. Interchange any two equations.

2. Multiply both sides of the equations by the same nonzero number.

3. Multiply any equation by a rumber, add the result to a second equation, and then replace the second equation with the sum.

 

To eliminate the x-term from the second equation, multiply both sides of the first equation by -1 and ass the result to the second equation.

 

 

 

 

Now use the same method to eirinale the x-term in the third equation.

Multely beth sides of the first equation by -1 and add the result to the third equation.

 

 

The equation system with the nea second and third equation is shown below. The next step is to eliminate the yeterm from the thad equation.

 

x + 6y - z = 19 

      y + 2z = 1 

-11y + 8z = 49

 

To eliminate the y-term in the third equation, multiply both sides of the second equation by 11 and add the result to the third equation.

 

 

Finally, to obtain an x-coefficient of 1 in the third equation, multiply both sides of the third equation by .

 

 

The system is now in echelon form. Solve by back-substitution.

x + 6y - z = -19 

        y + 2z = 1 

                z = 2

 

The third equation gives the value of z.

Back-substitute 2 for z in the second equation and solve for y.

 

y + 2(2) = 1 

y = -3

 

Finally, back-substitute the values of y and z in the first equation and solve for x.

 

             x + 6 (-3) - (2) = -19 

                      x = 1    

 

Therefore, the solution to the given system is as shown below.

          x = 1               y = -3              z = 2